$y'=0.5(100-y)$ Is $y=5e^{-0.5x}+100$ a solution to the above equation? Choose 1 answer: Choose 1 answer: (Choice A) A Yes (Choice B) B No
Answer: In order to find whether $y=5e^{-0.5x}+100$ is a solution, we need to substitute it into the equation and see if we get equivalent expressions on each side of the equal sign. In addition to substituting for $y$, we need to find the corresponding $y'$ expression to substitute into the equation: $\begin{aligned} y'&=\dfrac{d}{dx}\left[5e^{-0.5x}+100\right] \\\\ &=-2.5e^{-0.5x} \end{aligned}$ Now we substitute ${y=5e^{-0.5x}+100}$ and ${y'=-2.5e^{-0.5x}}$ into the equation: $\begin{aligned} {y'}&=0.5(100-{y}) \\\\ {-2.5e^{-0.5x}}&\stackrel{?}{=}0.5\left(100-\left({5e^{-0.5x}+100}\right)\right) \\\\ -2.5e^{-0.5x}&\stackrel{?}{=}50-2.5e^{-0.5x}-50 \\\\ -2.5e^{-0.5x}&\stackrel{\checkmark}{=}-2.5e^{-0.5x} \end{aligned}$ We obtained the same expression on each side. In conclusion, yes, $y=5e^{-0.5x}+100$ is a solution to the differential equation.